Question: $y^4+3xy+x=2$ Find the value of $\dfrac{dy}{dx}$ at the point $(2,0)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-2$ (Choice B) B $-\dfrac{1}{6}$ (Choice C) C $-6$ (Choice D) D $-\dfrac12$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $y^4+3xy+x=2$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} y^4+3xy+x&=2 \\\\ \dfrac{d}{dx}(y^4+3xy+x)&=\dfrac{d}{dx}(2) \\\\ \dfrac{d}{dx}(y^4)+\dfrac{d}{dx}(3xy)+\dfrac{d}{dx}(x)&=0 \\\\ 4y^3\cdot\dfrac{dy}{dx}+3\left(1\cdot y+x\cdot\dfrac{dy}{dx}\right)+1&=0 \\\\ 4y^3\cdot\dfrac{dy}{dx}+3y+3x\cdot\dfrac{dy}{dx}+1&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 4y^3\cdot\dfrac{dy}{dx}+3y+3x\cdot\dfrac{dy}{dx}+1&=0 \\\\ \dfrac{dy}{dx}(4y^3+3x)&=-(1+3y) \\\\ \dfrac{dy}{dx}&=-\dfrac{1+3y}{4y^3+3x} \end{aligned}$ Now we can plug the point $(2,0)$ into the expression for $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=-\dfrac{1+3y}{4y^3+3x} \\\\ &=-\dfrac{1+3(0)}{4(0)^3+3(2)} \gray{x=2,\,\,y=0} \\\\ &=-\dfrac{1}{6} \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at the point $(2,0)$ is $-\dfrac{1}{6}$.